Question

A bullet of mass m leaves a gun of mass M kept on a smooth horizontal surface. If the speed of the bullet relative to the gun is v, the recoil speed of the gun will be:

• A. $\frac{m}{M}$v
• B. $\frac{m}{M + m}$v
• C. $\frac{Mv}{M + m}$
• D. $\frac{M}{m}$v.

Answer 1

Answer: (B)

$\frac{m}{M + m}$v

Answer 2

No external force acts on the system (gun + bullet) during their impact (till the bullet leaves the gun). Therefore the momentum of the system remains constant. Before the impact the system (gun + bullet) was at rest. Hence its initial momentum is zero. Therefore just after the impact, its momentum of the system (gun + bullet) will be zero.

⇒ $m{\overrightarrow{v}}_{b} + M{\overrightarrow{v}}_{g}$ = 0

⇒ m $\left\lbrack {\overrightarrow{v}}_{bg} + {\overrightarrow{v}}_{g} \right\rbrack + M{\overrightarrow{v}}_{g}$ = 0

⇒ ${\overrightarrow{v}}_{g} = - \frac{m{\overrightarrow{v}}_{bg}}{M + m}$, where $\left| {\overrightarrow{v}}_{bg} \right|$

= Velocity of bullet relative gun = v

⇒ v_g = $\frac{mv}{M + m}$.

Opposite to the direction of v, Hence, the correct choice is